为什么这种分工导致零？ [英] Why does this division result in zero?

问题描述

我是C语言编写本code时，我遇到了以下问题。

I was writing this code in C when I encountered the following problem.

#include <stdio.h>
int main()
{
   int i=2;
   int j=3;
   int k,l;
   float a,b;
   k=i/j*j;
   l=j/i*i;
   a=i/j*j;
   b=j/i*i;
   printf("%d %d %f %f\n",k,l,a,b);
   return 0;
}

谁能告诉我为什么code是回零的第一个和第三个变量（ K 和 A ）？

推荐答案

我认为你遇到的是的整数运算的。您正确地假设→和 B 为2，但错误地认为 K 和 A 将在3，因为它是相同的操作。但它不是，它是整数运算（而不是浮点运算）。所以，当你做 I / J （请考虑使用一些空白），2/3 = 0.33333 ...的这是强制转换为 INT 的，从而变为0。然后，我们乘以3一遍，0 * 3 = 0。

What I think you are experiencing is integer arithmetic. You correctly suppose l and b to be 2, but incorrectly assume that k and a will be 3 because it's the same operation. But it's not, it's integer arithmetic (rather than floating-point arithmetic). So when you do i / j (please consider using some whitespace), 2 / 3 = 0.33333... which is cast to an int and thus becomes 0. Then we multiply by 3 again, and 0 * 3 = 0.

如果您更改 I 和Ĵ是浮动 S（或花椒你的数学与（浮点）蒙上），这会做你期望的。

If you change i and j to be floats (or pepper your math with (float) casts), this will do what you expect.

这篇关于为什么这种分工导致零？的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！